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# Systems of Linear Equations¶

A $$\textbf{linear equation}$$ is an equality of the form $$$\sum_{i = 1}^{n} (a_i x_i) = y,$$$$where$$a_i$$are scalars,$$x_i$$are unknown variables in$$\mathbb{R}$$, and$$y$ is a scalar.

TRY IT! Determine which of the following equations is linear and which is not. For the ones that are not linear, can you manipulate them so that they are?

1. $$3x_1 + 4x_2 - 3 = -5x_3$$

2. $$\frac{-x_1 + x_2}{x_3} = 2$$

3. $$x_1x_2 + x_3 = 5$$

Equation 1 can be rearranged to be $$3x_1 + 4x_2 + 5x_3= 3$$, which clearly has the form of a linear equation. Equation 2 is not linear but can be rearranged to be $$-x_1 + x_2 - 2x_3 = 0$$, which is linear. Equation 3 is not linear.

A $$\textbf{system of linear equations}$$ is a set of linear equations that share the same variables. Consider the following system of linear equations:

$\begin{eqnarray*} \begin{array}{rcrcccccrcc} a_{1,1} x_1 &+& a_{1,2} x_2 &+& {\ldots}& +& a_{1,n-1} x_{n-1} &+&a_{1,n} x_n &=& y_1,\\ a_{2,1} x_1 &+& a_{2,2} x_2 &+&{\ldots}& +& a_{2,n-1} x_{n-1} &+& a_{2,n} x_n &=& y_2, \\ &&&&{\ldots} &&{\ldots}&&&& \\ a_{m-1,1}x_1 &+& a_{m-1,2}x_2&+ &{\ldots}& +& a_{m-1,n-1} x_{n-1} &+& a_{m-1,n} x_n &=& y_{m-1},\\ a_{m,1} x_1 &+& a_{m,2}x_2 &+ &{\ldots}& +& a_{m,n-1} x_{n-1} &+& a_{m,n} x_n &=& y_{m}. \end{array} \end{eqnarray*}$

where $$a_{i,j}$$ and $$y_i$$ are real numbers. The $$\textbf{matrix form}$$ of a system of linear equations is $$\textbf{$$Ax = y$$}$$ where $$A$$ is a $${m} \times {n}$$ matrix, $$A(i,j) = a_{i,j}, y$$ is a vector in $${\mathbb{R}}^m$$, and $$x$$ is an unknown vector in $${\mathbb{R}}^n$$. The matrix form is showing as below:

$\begin{split}\begin{bmatrix} a_{1,1} & a_{1,2} & ... & a_{1,n}\\ a_{2,1} & a_{2,2} & ... & a_{2,n}\\ ... & ... & ... & ... \\ a_{m,1} & a_{m,2} & ... & a_{m,n} \end{bmatrix}\left[\begin{array}{c} x_1 \\x_2 \\ ... \\x_n \end{array}\right] = \left[\begin{array}{c} y_1 \\y_2 \\ ... \\y_m \end{array}\right]\end{split}$

If you carry out the matrix multiplication, you will see that you arrive back at the original system of equations.

TRY IT! Put the following system of equations into matrix form.

$\begin{eqnarray*} 4x + 3y - 5z &=& 2 \\ -2x - 4y + 5z &=& 5 \\ 7x + 8y &=& -3 \\ x + 2z &=& 1 \\ 9 + y - 6z &=& 6 \\ \end{eqnarray*}$
$\begin{split}\begin{bmatrix} 4 & 3 & -5\\ -2 & -4 & 5\\ 7 & 8 & 0\\ 1 & 0 & 2\\ 9 & 1 & -6 \end{bmatrix}\left[\begin{array}{c} x \\y \\z \end{array}\right] = \left[\begin{array}{c} 2 \\5 \\-3 \\1 \\6 \end{array}\right]\end{split}$